Question

if the digonals of a parallelogram are equal then show that it is a rectangle​

Answer 1

Answer:

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of

its interior angles is 90°

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

By SSS congruence rule,

ΔABC ≅ ΔDCB

So, ∠ABC = ∠DCB

It is known that the sum of measures of angles on the same side of traversal is 1800

∠ABC + ∠DCB = 180° [AB || CD]

=> ∠ABC + ∠ABC = 180°

=> 2∠ABC = 180°

=> ∠ABC = 90°

Since ABCD is a parallelogram and one of its interior angles is 90° ABCD is a rectangle.

Answer 2

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